$\frac{\mathrm{GMm}}{\mathrm{R}}+\mathrm{KE}=\frac{-\mathrm{GMm}}{3 \mathrm{R}}+\frac{1}{2} \mathrm{mV}^{2} \ldots(\mathrm{i})$ …(i) Get all questions and answers of Gravitation of NEET Physics on TopperLearning. $(1)-\frac{6 \mathrm{Gm}}{\mathrm{r}}$ It is to be launched from the earth’s surface out into free space. | A satellite is reolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h << R). Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. | NEET Previous Years Papers: greater than 10 lakh students appear for NEET every 12 months for almost 1 lakh seats in one-of-a-kind scientific and dental colleges across the us of a. | [AIEEE-2011] Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. $\frac{\mathrm{GMm}}{(3 \mathrm{R})^{2}}=\frac{\mathrm{mv}^{2}}{3 \mathrm{R}} \Rightarrow \mathrm{V}^{2}=\frac{\mathrm{GM}}{3 \mathrm{R}}$ ………..(ii) (4) $6.4 \times 10^{9}$ Joules | NEET Previous Year Question Paper for the year 2020, 2019, 2018, 2017, 2016 and 2015 along with answers and solutions in PDF format to download on Vedantu.com. The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere). (1) $\frac{5 \mathrm{GmM}}{6 \mathrm{R}}$         (2) $\frac{2 \mathrm{GmM}}{3 \mathrm{R}}$        (3) $\frac{\mathrm{GmM}}{2 \mathrm{R}}$        (4) $\frac{\mathrm{GmM}}{3 \mathrm{R}}$ $\mathrm{PE}_{\mathrm{i}}+\mathrm{KE}_{\mathrm{i}}=\mathrm{PE}_{\mathrm{f}}+\mathrm{KE}_{\mathrm{f}}$ For the convenience of candidates, we have provided more than 200 PDFs of Previous Year Papers. Taking gravitational potential V = 0 at r = $\infty$, the potential at the centre of the cavity thus formed is : (G = gravitational constant) Chapter-wise questions give … (1) $\frac{-2 \mathrm{GM}}{3 \mathrm{R}}$ Ncert Solutions Multiple Choice Questions (MCQ I) Hindi Physics. [JEE-Mains 2015] Articles Physics problems of NEET are comparatively easier. SHOW SOLUTION Download eSaral app for free study material and video tutorials. Past 40 Years Question Papers Solutions for NEET/AIIMS Physics Gravitation are provided here with simple step-by-step explanations. Gurukul helps the students in clearing and understanding each topic in a better way. (1) $6.4 \times 10^{10}$ Joules SHOW SOLUTION By principle of superosition | For JEE Main other Engineering Entrance Exam Preparation, JEE Main Physics Gravitation Previous Year Questions with Solutions is given below. [JEE-Mains 2013] from (i) & (ii) Amazing Facts That announces a lot approximately the type of opposition that medical aspirants have to go through as they begin their NEET … $(4)-\frac{4 \mathrm{Gm}}{\mathrm{r}}$ Notes SHOW SOLUTION Download India's Best Exam Preparation App. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas. option (2), $\mathrm{g}=\frac{\mathrm{GMx}}{\mathrm{R}^{3}}$ inside the Earth (straight line), $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ outside the Earth. (3) $\frac{-\mathrm{GM}}{2 \mathrm{R}}$ The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth’s radius) :-, Heat Transfer – JEE Main Previous Year Questions with Solutions, JEE Main Previous Year Questions Topicwise. where M is Mass of Earth Current Affairs Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas. Previous Year Show All; Question Type Show All; Difficulty Level Show All; Bookmarks; ... Gravitation; Mechanical Properties of Solids; Mechanical Properties of Fluids; $\mathrm{g}=\frac{\mathrm{GMx}}{\mathrm{R}^{3}}$ inside the Earth (straight line) | $\mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}^{3}}\left[3 \mathrm{R}^{2}-\frac{\mathrm{R}^{2}}{4}\right]+\frac{3 \mathrm{G}}{2} \frac{\mathrm{M}}{8 \frac{\mathrm{R}}{2}}$ 4M watch mins. The acceleration of the moon just before it strikes the earth in the previous question is Purchase Courses **This app is completely free!!! Practice Now. Communication System. Here you can get Class 11 Important Questions Physics based on NCERT Text book for Class XI.Physics Class 11 Important Questions are very helpful to score high marks in board exams. Media (2) Practice Now. Rigid Body Dynamics. $\mathrm{V}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ or $\sqrt{\mathrm{gR}}$ | $\Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=3 \Rightarrow \mathrm{h}=2 \mathrm{R}$, (1) $\frac{\mathrm{R}}        {2}$ (2) $\sqrt{2} \mathrm{R}$           (3) 2R             (4) $\frac{\mathrm{R}}{\sqrt{2}}$, $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}} \Rightarrow\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}=\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=9$, $\Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=3 \Rightarrow \mathrm{h}=2 \mathrm{R}$, Q. (2), Q. Free Download 31 Years NEET-AIPMT Chapter wise Solutions Physics. NEET Eligibility Criteria – Number of Attempts (2) Find PDF solutions, video solutions for NEET 2020 preparations. [AIEEE-2012] (1) Center Of Mass, Impulse. $\mathrm{V}_{\mathrm{e}} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ or $\sqrt{2 \mathrm{gR}}$ NEET questions & solutions with PDF and difficulty level. work done $=6.4 \times 10^{10} \mathrm{J}$, $\mathrm{PE}_{\mathrm{i}}+\mathrm{KE}_{\mathrm{i}}=\mathrm{PE}_{\mathrm{f}}+\mathrm{KE}_{\mathrm{f}}$, $-\mathrm{mgR}+\mathrm{KE}_{\mathrm{i}}=0+0$, $\mathrm{KE}_{\mathrm{i}}=+\mathrm{mgR}=1000 \times 10 \times 6.4 \times 10^{6}$, work done $=6.4 \times 10^{10} \mathrm{J}$, Q. SHOW SOLUTION we want latest years questions but it helps a lot, All topics are not covered but still nice collection, Thanks for this kind of helping hand.Waiting for more♥, Thanks for your collection, more questions would be great, Download India's Leading JEE | NEET | Class 9,10 Exam preparation app, Gravitation – JEE Main Previous Year Questions with Solutions, The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :-, Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is :-, Two particles of equal mass ‘m’ go around a circle of radius R under the action of their mutual gravitational attraction. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (3) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ Candidates who have appeared for class 12 more than one time for improvement can mention the year of the exam with a better score. Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET … Exercise well for Physics class 11 chapter 11 Gravitation with explanatory concept video solutions. The required energy for this work will be :- Graphical Questions. The value of ‘g’ and ‘R’ (radius of earth) are 10 m/s2 and 6400 km respectively. The speed of each particle is : From a solid sphere of mass M and radius R, a spherical portion of radius $\frac{\mathrm{R}}{2}$ is removed, as shown in the figure. 3 sec, OTP has been sent to your mobile number and is valid for one hour. The speed of each particle is : Also, NEET Previous Year Question Papers with solutions, online test series, model papers, NCERT Books, extra preparation books are present to help you prepare for NEET. $=\frac{-11 \mathrm{GM}}{8 \mathrm{R}}+\frac{3 \mathrm{GM}}{8 \mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}}$, (1) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$, (2) $\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$, (3) $\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$, (4) $\sqrt{2 \sqrt{2} \frac{\mathrm{GM}}{\mathrm{R}}}$, $\mathrm{F}_{\mathrm{net}}=\mathrm{F}_{1}+2 \mathrm{F}_{2} \cos 45^{\circ}=$ Centripetal force, $\Rightarrow \frac{\mathrm{GM}^{2}}{(2 \mathrm{R})^{2}}+\left[\frac{2 \mathrm{GM}^{2}}{(\sqrt{2} \mathrm{R})^{2}} \cos 45^{\circ}\right]=\frac{\mathrm{MV}^{2}}{\mathrm{R}}$, $\mathrm{V}=\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$, $\mathrm{V}=-\frac{\mathrm{GM}}{2 \mathrm{R}^{3}}\left[3 \mathrm{R}^{2}-\frac{\mathrm{R}^{2}}{4}\right]+\frac{3 \mathrm{G}}{2} \frac{\mathrm{M}}{8 \frac{\mathrm{R}}{2}}$, $=\frac{-11 \mathrm{GM}}{8 \mathrm{R}}+\frac{3 \mathrm{GM}}{8 \mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{R}}$, Q. Spend time answering previous years’ question papers. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Follow Us: Popular Chapters by Class: Class 6. The benefit of solving previous year papers of NEET (AIPMT) is that the students get to know the type of questions asked in the NEET exam and can have the experience of giving a real exam while solving the past year question paper of NEET. $\mathrm{F}_{\mathrm{net}}=\mathrm{F}_{1}+2 \mathrm{F}_{2} \cos 45^{\circ}=$ Centripetal force $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ outside the Earth (2), (1) $\sqrt{\frac{\mathrm{Gm}}{\mathrm{R}}}$         (2) $\sqrt{\frac{\mathrm{Gm}}{4 \mathrm{R}}}$         (3) $\sqrt{\frac{\mathrm{Gm}}{3 \mathrm{R}}}$           (4) $\sqrt{\frac{\mathrm{Gm}}{2 \mathrm{R}}}$, Q. Critical Thinking. 2017, 2016, 2015, 2014 chapter-wise Notes of Class 11th and 12th for... 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